23 votes 23 votes Consider the following statements: S1: The sum of two singular $n \times n$ matrices may be non-singular S2: The sum of two $n \times n$ non-singular matrices may be singular Which one of the following statements is correct? $S1$ and $S2$ both are true $S1$ is true, $S2$ is false $S1$ is false, $S2$ is true $S1$ and $S2$ both are false Linear Algebra gatecse-2001 linear-algebra normal matrix + – Kathleen asked Sep 14, 2014 edited Jun 7, 2018 by Milicevic3306 Kathleen 8.3k views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply Vishal Rana commented Aug 25, 2018 reply Follow Share I think that without taking any example we can get the answer. Following is the explaination for the two statements. statement 1 : let us consider any two matrices of n-by-n which are singular, where n >= 2 (beacuse for n = 1 we can only have 1 matrix which is singular). Now, the addition of these two matrices may be singular or nonsingular. Assume that all such possible pairs of matrices yield the sum as non-singular, then certainly statement 1 is true. Now, assume only some possible pairs of n-by-n matrices yield the sum as non-singular while others as singular, then also statement 1 is true, due to the keyword "may be" in the statement. Now, assume the last possibility that all pairs of such matrices yield as singular matrix , then also statement 1 is true again due to "may be" keyword. Hence, statement 1 is true for all cases. statement 2 : we can argue similarly that statement 2 is also true in all cases. Hence option A is right answer. Please correct me if I am wrong. 0 votes 0 votes KUSHAGRA गुप्ता commented Oct 14, 2019 reply Follow Share http://www.cse.iitd.ernet.in/~mittal/gate/gate_math_2001.html 6 votes 6 votes Kiyoshi commented Jan 4, 2022 reply Follow Share Both statements are possible.S1 : $A= \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}$ $B= \begin{bmatrix} 0 & 0 \\ 0 & 2 \end{bmatrix}$$ (A+B) = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$S2 : $A= \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ $B= \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}$$ (A+B) = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ 7 votes 7 votes Please log in or register to add a comment.
26 votes 26 votes A singular matrix is a square matrix whose determinant is 0. Whenever you solve question related to determinant, think of triangular matrix, like upper triangular matrix, because its determinant is just product of diagonal entries, and so easy to visualize. So definitely, in a singular matrix, one of the entries in diagonal must be zero. Similarly, in non-singular, none of the entries should be zero. Now take any matrices and just check . So we see that both S1 and S2 are true. So option (A) is correct. dragonball answered Apr 15, 2017 edited Apr 15, 2017 by dragonball dragonball comment Share Follow See 1 comment See all 1 1 comment reply Shiva Sagar Rao commented Feb 25, 2021 reply Follow Share This answer has been copied from here. When you copy, paste information from somewhere add source in your answer. 2 votes 2 votes Please log in or register to add a comment.
9 votes 9 votes ans should be A. for S1 : singular matrices matrix[0,0,0,1] + matrix[1,0,0,0] = matrix[1,0,0,1], which is non singular for S2 : non-singular matrices matrix[1,0,0,1] + matrix[0,1,1,0] = matrix[1,1,1,1], which is singular jayendra answered Dec 30, 2014 jayendra comment Share Follow See all 3 Comments See all 3 3 Comments reply skrahul commented Jan 17, 2016 reply Follow Share but matrices should be square ..? 0 votes 0 votes srestha commented Sep 2, 2016 reply Follow Share yes, that is true 0 votes 0 votes Verma Ashish commented Feb 4, 2019 i edited by Verma Ashish Sep 9, 2019 reply Follow Share They are square matrices-- $\begin{matrix} 0&0 \\ 0&1\end{matrix}\quad +$ $\quad\begin{matrix} 1&0 \\ 0&0\end{matrix}$ $=\begin{matrix} 1&0\\0&1\end{matrix}$ $\Rightarrow (singular+singular)\;$ may be $non\; singular$ 4 votes 4 votes Please log in or register to add a comment.
5 votes 5 votes det (A + B) is not equal to det (A) + det (B), it means that if A and B both are singular then, the determinant of sum of A and B may not to be equal to 0. S1 is correct Similarly, if A and B are non singular matrix, then det(A+B) may not be Non Zero, it can be Zero means Singular. S2 is also correct Aakash_ answered Jun 23, 2018 Aakash_ comment Share Follow See all 0 reply Please log in or register to add a comment.
3 votes 3 votes I think there is a contradiction Let me explain with example If A = [ 2 10 1 5] B = [ 3 6 2 4] Sum [ 5 16 3 9] | sum| = 9×5_16×3=- 3 which is not equal to zero . So it is non singular signals8692 answered Nov 18, 2016 edited Dec 28, 2016 by Vijay Thakur signals8692 comment Share Follow See 1 comment See all 1 1 comment reply nocturnal123 commented Oct 1, 2020 reply Follow Share I too, took a specific example and found that “ The sum of two singular n×n matrices was resulting into singular matrix ”. I think, this is not contradiction to given statement “ The sum of two singular n×n matrices may be non-singular ”. If the statement would have been given as “ The sum of two singular n×n matrices must be non-singular ”, then contradiction would have been present and hence this statement would be marked as false. 0 votes 0 votes Please log in or register to add a comment.